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Calculating Average Value with Fundamental Theorem of Calculus

Created by @nathanedwards
 at November 1st 2023, 12:18:07 am.
AP_Calculus_AB-Questions
#calculus
#integrals
#average-value

Question:

Consider the function

f(x)=∫0x(6t2+4t−5) dt.f(x) = \int_{0}^{x} (6t^2 + 4t - 5) \ dt.(a)

(b) Find the exact value of f(3).

(c) Determine the average value of f(x) on the interval [0, 3].

Answer:

(a) According to the Fundamental Theorem of Calculus, if f(x) is a continuous function on [a, b] and F(x) is an antiderivative of f(x) on [a, b], then:

∫abf(x) dx=F(b)−F(a).\int_{a}^{b} f(x) \ dx = F(b) - F(a).

Let's find an antiderivative of (6t^2 + 4t - 5) first.

F(t)=∫(6t2+4t−5) dt=2t3+2t2−5t+CF(t) = \int (6t^2 + 4t - 5) \ dt = 2t^3 + 2t^2 - 5t + C

Now, we can make use of the Fundamental Theorem of Calculus to find f(x):

f(x)=F(x)−F(0)=2x3+2x2−5x+C−C=2x3+2x2−5xf(x) = F(x) - F(0) = 2x^3 + 2x^2 - 5x + C - C = 2x^3 + 2x^2 - 5x

Thus, the equation for f(x) is f(x) = 2x^3 + 2x^2 - 5x.

(b) To find f(3), we substitute x = 3 into the equation for f(x).

f(3)=2(3)3+2(3)2−5(3)=2(27)+2(9)−15=54+18−15=57f(3) = 2(3)^3 + 2(3)^2 - 5(3) = 2(27) + 2(9) - 15 = 54 + 18 - 15 = 57

Therefore, f(3) = 57.

(c) The average value of f(x) on the interval [0, 3] is given by:

1b−a∫abf(x) dx\frac{1}{b - a} \int_{a}^{b} f(x) \ dx

Substituting the interval limits [a, b] = [0, 3]:

13−0∫03f(x) dx=13∫03(2x3+2x2−5x) dx=13[12x4+23x3−52x2]∣03=13[12(3)4+23(3)3−52(3)2]−13(0)=13[12(81)+23(27)−52(9)]=13[812+543−452]=13[812+18−452]=13[812−452+18]=13[81−452+18]=13[362+18]=13[18+18]=13⋅36=12\frac{1}{3 - 0} \int_{0}^{3} f(x) \ dx = \frac{1}{3} \int_{0}^{3} (2x^3 + 2x^2 - 5x) \ dx = \frac{1}{3} \bigg[\frac{1}{2}x^4 + \frac{2}{3}x^3 - \frac{5}{2}x^2\bigg] \bigg|_{0}^{3} = \frac{1}{3} \bigg[\frac{1}{2}(3)^4 + \frac{2}{3}(3)^3 - \frac{5}{2}(3)^2\bigg] - \frac{1}{3}(0) = \frac{1}{3} \bigg[\frac{1}{2}(81) + \frac{2}{3}(27) - \frac{5}{2}(9)\bigg] = \frac{1}{3} \bigg[\frac{81}{2} + \frac{54}{3} - \frac{45}{2}\bigg] = \frac{1}{3} \bigg[\frac{81}{2} + 18 - \frac{45}{2}\bigg] = \frac{1}{3} \bigg[\frac{81}{2} - \frac{45}{2} + 18\bigg] = \frac{1}{3} \bigg[\frac{81 - 45}{2} + 18\bigg] = \frac{1}{3} \bigg[\frac{36}{2} + 18\bigg] = \frac{1}{3} \bigg[18 + 18\bigg] = \frac{1}{3} \cdot 36 = 12

Therefore, the average value of f(x) on the interval [0, 3] is 12.

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